3.898 \(\int \frac {\tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=154 \[ \frac {\tan ^{10}(c+d x)}{10 a d}+\frac {63 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac {\tan ^9(c+d x) \sec (c+d x)}{10 a d}+\frac {9 \tan ^7(c+d x) \sec (c+d x)}{80 a d}-\frac {21 \tan ^5(c+d x) \sec (c+d x)}{160 a d}+\frac {21 \tan ^3(c+d x) \sec (c+d x)}{128 a d}-\frac {63 \tan (c+d x) \sec (c+d x)}{256 a d} \]

[Out]

63/256*arctanh(sin(d*x+c))/a/d-63/256*sec(d*x+c)*tan(d*x+c)/a/d+21/128*sec(d*x+c)*tan(d*x+c)^3/a/d-21/160*sec(
d*x+c)*tan(d*x+c)^5/a/d+9/80*sec(d*x+c)*tan(d*x+c)^7/a/d-1/10*sec(d*x+c)*tan(d*x+c)^9/a/d+1/10*tan(d*x+c)^10/a
/d

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Rubi [A]  time = 0.19, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2706, 2607, 30, 2611, 3770} \[ \frac {\tan ^{10}(c+d x)}{10 a d}+\frac {63 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac {\tan ^9(c+d x) \sec (c+d x)}{10 a d}+\frac {9 \tan ^7(c+d x) \sec (c+d x)}{80 a d}-\frac {21 \tan ^5(c+d x) \sec (c+d x)}{160 a d}+\frac {21 \tan ^3(c+d x) \sec (c+d x)}{128 a d}-\frac {63 \tan (c+d x) \sec (c+d x)}{256 a d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^9/(a + a*Sin[c + d*x]),x]

[Out]

(63*ArcTanh[Sin[c + d*x]])/(256*a*d) - (63*Sec[c + d*x]*Tan[c + d*x])/(256*a*d) + (21*Sec[c + d*x]*Tan[c + d*x
]^3)/(128*a*d) - (21*Sec[c + d*x]*Tan[c + d*x]^5)/(160*a*d) + (9*Sec[c + d*x]*Tan[c + d*x]^7)/(80*a*d) - (Sec[
c + d*x]*Tan[c + d*x]^9)/(10*a*d) + Tan[c + d*x]^10/(10*a*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int \sec ^2(c+d x) \tan ^9(c+d x) \, dx}{a}-\frac {\int \sec (c+d x) \tan ^{10}(c+d x) \, dx}{a}\\ &=-\frac {\sec (c+d x) \tan ^9(c+d x)}{10 a d}+\frac {9 \int \sec (c+d x) \tan ^8(c+d x) \, dx}{10 a}+\frac {\operatorname {Subst}\left (\int x^9 \, dx,x,\tan (c+d x)\right )}{a d}\\ &=\frac {9 \sec (c+d x) \tan ^7(c+d x)}{80 a d}-\frac {\sec (c+d x) \tan ^9(c+d x)}{10 a d}+\frac {\tan ^{10}(c+d x)}{10 a d}-\frac {63 \int \sec (c+d x) \tan ^6(c+d x) \, dx}{80 a}\\ &=-\frac {21 \sec (c+d x) \tan ^5(c+d x)}{160 a d}+\frac {9 \sec (c+d x) \tan ^7(c+d x)}{80 a d}-\frac {\sec (c+d x) \tan ^9(c+d x)}{10 a d}+\frac {\tan ^{10}(c+d x)}{10 a d}+\frac {21 \int \sec (c+d x) \tan ^4(c+d x) \, dx}{32 a}\\ &=\frac {21 \sec (c+d x) \tan ^3(c+d x)}{128 a d}-\frac {21 \sec (c+d x) \tan ^5(c+d x)}{160 a d}+\frac {9 \sec (c+d x) \tan ^7(c+d x)}{80 a d}-\frac {\sec (c+d x) \tan ^9(c+d x)}{10 a d}+\frac {\tan ^{10}(c+d x)}{10 a d}-\frac {63 \int \sec (c+d x) \tan ^2(c+d x) \, dx}{128 a}\\ &=-\frac {63 \sec (c+d x) \tan (c+d x)}{256 a d}+\frac {21 \sec (c+d x) \tan ^3(c+d x)}{128 a d}-\frac {21 \sec (c+d x) \tan ^5(c+d x)}{160 a d}+\frac {9 \sec (c+d x) \tan ^7(c+d x)}{80 a d}-\frac {\sec (c+d x) \tan ^9(c+d x)}{10 a d}+\frac {\tan ^{10}(c+d x)}{10 a d}+\frac {63 \int \sec (c+d x) \, dx}{256 a}\\ &=\frac {63 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac {63 \sec (c+d x) \tan (c+d x)}{256 a d}+\frac {21 \sec (c+d x) \tan ^3(c+d x)}{128 a d}-\frac {21 \sec (c+d x) \tan ^5(c+d x)}{160 a d}+\frac {9 \sec (c+d x) \tan ^7(c+d x)}{80 a d}-\frac {\sec (c+d x) \tan ^9(c+d x)}{10 a d}+\frac {\tan ^{10}(c+d x)}{10 a d}\\ \end {align*}

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Mathematica [A]  time = 2.42, size = 122, normalized size = 0.79 \[ \frac {\frac {2 \left (965 \sin ^8(c+d x)+325 \sin ^7(c+d x)-2045 \sin ^6(c+d x)-765 \sin ^5(c+d x)+1923 \sin ^4(c+d x)+643 \sin ^3(c+d x)-827 \sin ^2(c+d x)-187 \sin (c+d x)+128\right )}{(\sin (c+d x)-1)^4 (\sin (c+d x)+1)^5}+630 \tanh ^{-1}(\sin (c+d x))}{2560 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^9/(a + a*Sin[c + d*x]),x]

[Out]

(630*ArcTanh[Sin[c + d*x]] + (2*(128 - 187*Sin[c + d*x] - 827*Sin[c + d*x]^2 + 643*Sin[c + d*x]^3 + 1923*Sin[c
+ d*x]^4 - 765*Sin[c + d*x]^5 - 2045*Sin[c + d*x]^6 + 325*Sin[c + d*x]^7 + 965*Sin[c + d*x]^8))/((-1 + Sin[c
+ d*x])^4*(1 + Sin[c + d*x])^5))/(2560*a*d)

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fricas [A]  time = 0.53, size = 187, normalized size = 1.21 \[ \frac {1930 \, \cos \left (d x + c\right )^{8} - 3630 \, \cos \left (d x + c\right )^{6} + 3156 \, \cos \left (d x + c\right )^{4} - 1488 \, \cos \left (d x + c\right )^{2} + 315 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 315 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (325 \, \cos \left (d x + c\right )^{6} - 210 \, \cos \left (d x + c\right )^{4} + 88 \, \cos \left (d x + c\right )^{2} - 16\right )} \sin \left (d x + c\right ) + 288}{2560 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^9/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/2560*(1930*cos(d*x + c)^8 - 3630*cos(d*x + c)^6 + 3156*cos(d*x + c)^4 - 1488*cos(d*x + c)^2 + 315*(cos(d*x +
c)^8*sin(d*x + c) + cos(d*x + c)^8)*log(sin(d*x + c) + 1) - 315*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8
)*log(-sin(d*x + c) + 1) - 2*(325*cos(d*x + c)^6 - 210*cos(d*x + c)^4 + 88*cos(d*x + c)^2 - 16)*sin(d*x + c) +
288)/(a*d*cos(d*x + c)^8*sin(d*x + c) + a*d*cos(d*x + c)^8)

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giac [A]  time = 0.37, size = 156, normalized size = 1.01 \[ \frac {\frac {1260 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {1260 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {5 \, {\left (525 \, \sin \left (d x + c\right )^{4} - 1580 \, \sin \left (d x + c\right )^{3} + 1818 \, \sin \left (d x + c\right )^{2} - 932 \, \sin \left (d x + c\right ) + 177\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac {2877 \, \sin \left (d x + c\right )^{5} + 9265 \, \sin \left (d x + c\right )^{4} + 12030 \, \sin \left (d x + c\right )^{3} + 7430 \, \sin \left (d x + c\right )^{2} + 1965 \, \sin \left (d x + c\right ) + 113}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{10240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^9/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/10240*(1260*log(abs(sin(d*x + c) + 1))/a - 1260*log(abs(sin(d*x + c) - 1))/a + 5*(525*sin(d*x + c)^4 - 1580*
sin(d*x + c)^3 + 1818*sin(d*x + c)^2 - 932*sin(d*x + c) + 177)/(a*(sin(d*x + c) - 1)^4) - (2877*sin(d*x + c)^5
+ 9265*sin(d*x + c)^4 + 12030*sin(d*x + c)^3 + 7430*sin(d*x + c)^2 + 1965*sin(d*x + c) + 113)/(a*(sin(d*x + c
) + 1)^5))/d

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maple [A]  time = 0.46, size = 198, normalized size = 1.29 \[ \frac {1}{256 a d \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {1}{32 a d \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {57}{512 a d \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {65}{256 a d \left (\sin \left (d x +c \right )-1\right )}-\frac {63 \ln \left (\sin \left (d x +c \right )-1\right )}{512 a d}+\frac {1}{160 a d \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {13}{256 a d \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {23}{128 a d \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {187}{512 a d \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{2 a d \left (1+\sin \left (d x +c \right )\right )}+\frac {63 \ln \left (1+\sin \left (d x +c \right )\right )}{512 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^9*sin(d*x+c)^9/(a+a*sin(d*x+c)),x)

[Out]

1/256/a/d/(sin(d*x+c)-1)^4+1/32/a/d/(sin(d*x+c)-1)^3+57/512/a/d/(sin(d*x+c)-1)^2+65/256/a/d/(sin(d*x+c)-1)-63/
512/a/d*ln(sin(d*x+c)-1)+1/160/a/d/(1+sin(d*x+c))^5-13/256/a/d/(1+sin(d*x+c))^4+23/128/a/d/(1+sin(d*x+c))^3-18
7/512/a/d/(1+sin(d*x+c))^2+1/2/a/d/(1+sin(d*x+c))+63/512*ln(1+sin(d*x+c))/a/d

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maxima [A]  time = 0.52, size = 214, normalized size = 1.39 \[ \frac {\frac {2 \, {\left (965 \, \sin \left (d x + c\right )^{8} + 325 \, \sin \left (d x + c\right )^{7} - 2045 \, \sin \left (d x + c\right )^{6} - 765 \, \sin \left (d x + c\right )^{5} + 1923 \, \sin \left (d x + c\right )^{4} + 643 \, \sin \left (d x + c\right )^{3} - 827 \, \sin \left (d x + c\right )^{2} - 187 \, \sin \left (d x + c\right ) + 128\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} + \frac {315 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {315 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{2560 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^9/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/2560*(2*(965*sin(d*x + c)^8 + 325*sin(d*x + c)^7 - 2045*sin(d*x + c)^6 - 765*sin(d*x + c)^5 + 1923*sin(d*x +
c)^4 + 643*sin(d*x + c)^3 - 827*sin(d*x + c)^2 - 187*sin(d*x + c) + 128)/(a*sin(d*x + c)^9 + a*sin(d*x + c)^8
- 4*a*sin(d*x + c)^7 - 4*a*sin(d*x + c)^6 + 6*a*sin(d*x + c)^5 + 6*a*sin(d*x + c)^4 - 4*a*sin(d*x + c)^3 - 4*
a*sin(d*x + c)^2 + a*sin(d*x + c) + a) + 315*log(sin(d*x + c) + 1)/a - 315*log(sin(d*x + c) - 1)/a)/d

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mupad [B]  time = 17.05, size = 497, normalized size = 3.23 \[ \frac {63\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{128\,a\,d}-\frac {\frac {63\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{128}+\frac {63\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}}{64}-\frac {105\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{32}-\frac {483\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{64}+\frac {1407\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{160}+\frac {8043\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{320}-\frac {1779\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{160}-\frac {15159\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{320}+\frac {245\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{64}-\frac {15159\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{320}-\frac {1779\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{160}+\frac {8043\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{320}+\frac {1407\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160}-\frac {483\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64}-\frac {105\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{32}+\frac {63\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{64}+\frac {63\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}-7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}-16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}+20\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+56\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-28\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-112\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+14\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+140\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+14\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-112\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-28\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+56\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+20\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^9/(cos(c + d*x)^9*(a + a*sin(c + d*x))),x)

[Out]

(63*atanh(tan(c/2 + (d*x)/2)))/(128*a*d) - ((63*tan(c/2 + (d*x)/2))/128 + (63*tan(c/2 + (d*x)/2)^2)/64 - (105*
tan(c/2 + (d*x)/2)^3)/32 - (483*tan(c/2 + (d*x)/2)^4)/64 + (1407*tan(c/2 + (d*x)/2)^5)/160 + (8043*tan(c/2 + (
d*x)/2)^6)/320 - (1779*tan(c/2 + (d*x)/2)^7)/160 - (15159*tan(c/2 + (d*x)/2)^8)/320 + (245*tan(c/2 + (d*x)/2)^
9)/64 - (15159*tan(c/2 + (d*x)/2)^10)/320 - (1779*tan(c/2 + (d*x)/2)^11)/160 + (8043*tan(c/2 + (d*x)/2)^12)/32
0 + (1407*tan(c/2 + (d*x)/2)^13)/160 - (483*tan(c/2 + (d*x)/2)^14)/64 - (105*tan(c/2 + (d*x)/2)^15)/32 + (63*t
an(c/2 + (d*x)/2)^16)/64 + (63*tan(c/2 + (d*x)/2)^17)/128)/(d*(a + 2*a*tan(c/2 + (d*x)/2) - 7*a*tan(c/2 + (d*x
)/2)^2 - 16*a*tan(c/2 + (d*x)/2)^3 + 20*a*tan(c/2 + (d*x)/2)^4 + 56*a*tan(c/2 + (d*x)/2)^5 - 28*a*tan(c/2 + (d
*x)/2)^6 - 112*a*tan(c/2 + (d*x)/2)^7 + 14*a*tan(c/2 + (d*x)/2)^8 + 140*a*tan(c/2 + (d*x)/2)^9 + 14*a*tan(c/2
+ (d*x)/2)^10 - 112*a*tan(c/2 + (d*x)/2)^11 - 28*a*tan(c/2 + (d*x)/2)^12 + 56*a*tan(c/2 + (d*x)/2)^13 + 20*a*t
an(c/2 + (d*x)/2)^14 - 16*a*tan(c/2 + (d*x)/2)^15 - 7*a*tan(c/2 + (d*x)/2)^16 + 2*a*tan(c/2 + (d*x)/2)^17 + a*
tan(c/2 + (d*x)/2)^18))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**9*sin(d*x+c)**9/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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